Complex exercise asked by Dan Sitaru in IMO-Do you speak mathematics ? group

Exercise:

Solve in $\mathbb{C},$ $\left| z-2016 \right|+\left| z-2015 \right|=1$

Solution: we know that $\left| {{z}_{1}}+{{z}_{2}} \right|\le \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|$ where ${{z}_{1}},{{z}_{2}}\in \mathbb{C}$

So $\left| z-2016 \right|+\left| z-2015 \right|\ge \left| z-2016+z-2015 \right|$

Hence $\left| 2z-4031 \right|\le 1$   (*)

Let $z=x+iy\,\,,\,\,\,\left( x,y \right)\in {{\mathbb{R}}^{2}}$

Now $\left| z-2016 \right|+\left| z-2015 \right|=1\Leftrightarrow \left| z-2016 \right|=1-\left| z-2015 \right|$ S.B.S

$\Rightarrow {{\left| z-2016 \right|}^{2}}=1+{{\left| z-2015 \right|}^{2}}-2\left| z-2015 \right|$

$\Rightarrow {{\left( x-2016 \right)}^{2}}+{{y}^{2}}=1+{{\left( x-2015 \right)}^{2}}+{{y}^{2}}-2\left| z-2015 \right|$

$\Rightarrow -\left( 2x-4031 \right)=1-2\sqrt{{{\left( x-2015 \right)}^{2}}+{{y}^{2}}}$

$\Rightarrow x-2015=\sqrt{{{\left( x-2015 \right)}^{2}}+{{y}^{2}}}$  S.B.S

$\Rightarrow {{\left( x-2015 \right)}^{2}}={{\left( x-2015 \right)}^{2}}+{{y}^{2}}\Leftrightarrow {{y}^{2}}=0\Leftrightarrow y=0$ so $\operatorname{Im}\left( z \right)=0$

Now using (*) $\left| 2z-4031 \right|\le 1\Leftrightarrow \left| 2x-4031 \right|\le 1\Leftrightarrow -1+4031\le 2x\le 1+4031\Leftrightarrow 2015\le x\le 2016$

Thus $2015\le \operatorname{Re}\left( z \right)\le 2016$